Question

$$\left. \begin{array} { l } { 2 x + 7 = 5 } \\ { 10 x + 5 = 10 } \end{array} \right.$$

Answer

$$r=(510*x+503)/(2*e^3*p^2*a*t*o*x)$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{e}^{3}{p}^{2}rato\times 2x+7=510x+510\]
Regroup terms.
\[2{e}^{3}{p}^{2}ratox+7=510x+510\]
Subtract \(7\) from both sides.
\[2{e}^{3}{p}^{2}ratox=510x+510-7\]
Simplify  \(510x+510-7\)  to  \(510x+503\).
\[2{e}^{3}{p}^{2}ratox=510x+503\]
Divide both sides by \(2\).
\[{e}^{3}{p}^{2}ratox=\frac{510x+503}{2}\]
Divide both sides by \({e}^{3}\).
\[{p}^{2}ratox=\frac{\frac{510x+503}{2}}{{e}^{3}}\]
Simplify  \(\frac{\frac{510x+503}{2}}{{e}^{3}}\)  to  \(\frac{510x+503}{2{e}^{3}}\).
\[{p}^{2}ratox=\frac{510x+503}{2{e}^{3}}\]
Divide both sides by \({p}^{2}\).
\[ratox=\frac{\frac{510x+503}{2{e}^{3}}}{{p}^{2}}\]
Simplify  \(\frac{\frac{510x+503}{2{e}^{3}}}{{p}^{2}}\)  to  \(\frac{510x+503}{2{e}^{3}{p}^{2}}\).
\[ratox=\frac{510x+503}{2{e}^{3}{p}^{2}}\]
Divide both sides by \(a\).
\[rtox=\frac{\frac{510x+503}{2{e}^{3}{p}^{2}}}{a}\]
Simplify  \(\frac{\frac{510x+503}{2{e}^{3}{p}^{2}}}{a}\)  to  \(\frac{510x+503}{2{e}^{3}{p}^{2}a}\).
\[rtox=\frac{510x+503}{2{e}^{3}{p}^{2}a}\]
Divide both sides by \(t\).
\[rox=\frac{\frac{510x+503}{2{e}^{3}{p}^{2}a}}{t}\]
Simplify  \(\frac{\frac{510x+503}{2{e}^{3}{p}^{2}a}}{t}\)  to  \(\frac{510x+503}{2{e}^{3}{p}^{2}at}\).
\[rox=\frac{510x+503}{2{e}^{3}{p}^{2}at}\]
Divide both sides by \(o\).
\[rx=\frac{\frac{510x+503}{2{e}^{3}{p}^{2}at}}{o}\]
Simplify  \(\frac{\frac{510x+503}{2{e}^{3}{p}^{2}at}}{o}\)  to  \(\frac{510x+503}{2{e}^{3}{p}^{2}ato}\).
\[rx=\frac{510x+503}{2{e}^{3}{p}^{2}ato}\]
Divide both sides by \(x\).
\[r=\frac{\frac{510x+503}{2{e}^{3}{p}^{2}ato}}{x}\]
Simplify  \(\frac{\frac{510x+503}{2{e}^{3}{p}^{2}ato}}{x}\)  to  \(\frac{510x+503}{2{e}^{3}{p}^{2}atox}\).
\[r=\frac{510x+503}{2{e}^{3}{p}^{2}atox}\]